Calculate the exrtreme value of the function f(x) = 3x^2-4x+5
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
At the extreme values of a function f(x), the first derivative f'(x) = 0
f(x) = 3x^2 - 4x + 5
f'(x) = 6x - 4
6x - 4 = 0
=> x = 4/6
=> x = 2/3
The extreme value of the function is f(2/3) = 3*4/9 - 4*2/3 + 5
=> 4/3 - 8/3 + 5
=> 11/3
The extreme value of the function is 11/3
hala718 | High School Teacher | (Level 1) Educator Emeritus
f(x)= 3x^2 - 4x + 5
we need to find the extreme value of f(x).
First we will find the critical points.
==. f'(x) = 6x -4 = 0
==> x = 4/6 = 2/3
Then the function f(x) has an extreme value at f(2/3)
==> f(2/3) = 3*(2/3)^2 - 4*2/3 + 5
==> f(2/3) = 12/9 - 8/3 + 5 = ( 12 - 24 + 45) / 9 = 33/9 = 11/3
Then the function has extreme value at the point f(2/3) = 11/3 OR the point ( 2/3, 11/3)